We have the function
$y=\sin ^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)$
For $y$ to be defined $\left|\frac{x^{2}}{1+x^{2}}\right|<1$ which is true for all $x \in R$
Now, $y=\sin ^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)$
$\Rightarrow \frac{x^{2}}{1+x^{2}}=\sin y$
$\Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}$
For the existance of $x$ $\sin y \geq 0$ and $1-\sin y>0$
$\Rightarrow 0 \leq \sin y<1$
$\Rightarrow 0 \leq y<\frac{\pi}{2}$
Thus, range of the given function is $\left[0, \frac{\pi}{2}\right)$.